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Question Number 201969 by BaliramKumar last updated on 17/Dec/23

$$\mathrm{A}\mathrm{dice}\mathrm{is}\mathrm{cast}\mathrm{twice},\mathrm{and}\mathrm{the}\mathrm{sum}$$$$\mathrm{of}\mathrm{the}\mathrm{appearing}\mathrm{numbers}\mathrm{is}10.$$$$\mathrm{The}\mathrm{probability}\mathrm{that}\mathrm{the}\mathrm{number}5\mathrm{has}$$$$\mathrm{appeared}\mathrm{at}\mathrm{least}\mathrm{once}\mathrm{is}.$$

Answered by Frix last updated on 17/Dec/23

$$3\mathrm{possible}\mathrm{cases}$$$$4+6$$$$5+5$$$$6+4$$$$\Rightarrow p=\frac{1}{3}$$

Commented by BaliramKumar last updated on 17/Dec/23

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by mr W last updated on 18/Dec/23

$$p=\frac{1}{6\times 6}=\frac{1}{36}$$

Commented by Frix last updated on 18/Dec/23

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{after}\mathrm{the}\mathrm{throws}\mathrm{and}\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{the}$$$$\mathrm{sum}\mathrm{is}10.$$

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